4 ms·

Suppose 5/0 = infinity. What does it buy you? At the very least, one would expect that 5/0 = infinity implies that 5 = 0 * infinity, but that is not the case.

by eipiman 18y ago

Suppose 5/0 = infinity. What does it buy you? At the very least, one would expect that 5/0 = infinity implies that 5 = 0 * infinity, but that is not the case. And if you insist that it is, does 4/0 = infinity then imply that 4 = 0 * infinity, and hence 4 = 5?

speek 18y ago
Infinity is not a number... its a concept. Say you were to count every integer, you would count to infinity. Now what if you were to count every number between 0 and 1? Would it still be infinity? I've always thought in degrees of infinity, but either way its still not a number. If you kept the degrees of infinity separate, you could divide 5 by zero and it would not be the same as dividing 4 by zero. If you wanted to play with limits, you could say the limit of 5/x as x approaches 0 is infinity, but we don't really have a name for what x/0 would be. Someone did come up with nullity, but that's not good enough. Plus, you can't have Zero of something. Having implies a positive net, thus it's not correct to say that you have zero of something... you could say you don't have something, but they're semantically different. Zero is just another concept.
- abstractbill 18y ago
  Take a look at Cantor's work, if you're interested in this stuff. The number of integers is known as a countable infinity. The number of numbers between 0 and 1 is called an uncountable infinity. The proof that these are not the same (Cantor's diagonalization argument) is my favorite proof - it's really neat. It's quite clear and maybe even obvious in retrospect (I like to explain it to friends who are curious about higher mathematics), but probably not something you would have come up with if you weren't Cantor! You can ask if there's an infinity between these two, and that's actually unprovable. You can prove that, without adding more axioms to mathematics, it's impossible to prove one way or another whether there's an infinity between them. Sometimes I really miss this stuff.
  - Retric 18y ago
    I never really accepted Cantor's idea. First off if you have a sequence ...1 and ...0 at infinity they are the same number so saying you flip the last bit does not demonstrate that the above number is not in your set. Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity. PS: This is why I went to CS there is no way to verify that a given proof or program is accurate but that does not stop a program from being useful. Edit: Restating the second argument: Start with all zero's, count in binary from the left, 1000.. > 0100.. > 1100.. > ... and the diagonal becomes 1111111 which is basically just a higher order of infinity like N = the sum of all prior numbers.
    - hhm 18y ago
      I never really accepted Cantor's idea. First off if you have a sequence ...1 and ...0 at infinity they are the same number so saying you flip the last bit does not demonstrate that the above number is not in your set. It doesn't work like that. You want to show that you can't enumerate all reals, so what do you do? You start enumerating all reals. Then you build a new number, whose Nth bit is different from the Nth bit of the Nth real number. You can do that because you have enumerated them, and we have never talked about an infinite N so we are never flipping the last bit of the last number nor anything similar. We are only flipping the Nth bit of the Nth number. Number 819439243? We flip the 819439243th bit. That's all we do. What do you do then? Well, you compare that new number you created, with the first number, the second number, and so on. And you see that this number is different to all the other numbers. What does it mean? That it doesn't matter what your enumeration scheme is, there is at least one number that will always be out of your enumeration scheme, and so your enumeration scheme will fail (so this is an absurd also, as we were supposing that this specific enumeration scheme -that could be any enum scheme- was complete). So you don't have as many integers as reals, so QED. Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity. As you see, it doens't matter, as in this proof you never need to do the sum of all numbers.
      - Retric 18y ago

by eipiman 18y ago

speek 18y ago

Infinity is not a number... its a concept. Say you were to count every integer, you would count to infinity. Now what if you were to count every number between 0 and 1? Would it still be infinity? I've always thought in degrees of infinity, but either way its still not a number. If you kept the degrees of infinity separate, you could divide 5 by zero and it would not be the same as dividing 4 by zero. If you wanted to play with limits, you could say the limit of 5/x as x approaches 0 is infinity, but we don't really have a name for what x/0 would be. Someone did come up with nullity, but that's not good enough. Plus, you can't have Zero of something. Having implies a positive net, thus it's not correct to say that you have zero of something... you could say you don't have something, but they're semantically different. Zero is just another concept.

abstractbill 18y ago
Take a look at Cantor's work, if you're interested in this stuff. The number of integers is known as a countable infinity. The number of numbers between 0 and 1 is called an uncountable infinity. The proof that these are not the same (Cantor's diagonalization argument) is my favorite proof - it's really neat. It's quite clear and maybe even obvious in retrospect (I like to explain it to friends who are curious about higher mathematics), but probably not something you would have come up with if you weren't Cantor! You can ask if there's an infinity between these two, and that's actually unprovable. You can prove that, without adding more axioms to mathematics, it's impossible to prove one way or another whether there's an infinity between them. Sometimes I really miss this stuff.
- Retric 18y ago
  I never really accepted Cantor's idea. First off if you have a sequence ...1 and ...0 at infinity they are the same number so saying you flip the last bit does not demonstrate that the above number is not in your set. Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity. PS: This is why I went to CS there is no way to verify that a given proof or program is accurate but that does not stop a program from being useful. Edit: Restating the second argument: Start with all zero's, count in binary from the left, 1000.. > 0100.. > 1100.. > ... and the diagonal becomes 1111111 which is basically just a higher order of infinity like N = the sum of all prior numbers.
  - hhm 18y ago
    I never really accepted Cantor's idea. First off if you have a sequence ...1 and ...0 at infinity they are the same number so saying you flip the last bit does not demonstrate that the above number is not in your set. It doesn't work like that. You want to show that you can't enumerate all reals, so what do you do? You start enumerating all reals. Then you build a new number, whose Nth bit is different from the Nth bit of the Nth real number. You can do that because you have enumerated them, and we have never talked about an infinite N so we are never flipping the last bit of the last number nor anything similar. We are only flipping the Nth bit of the Nth number. Number 819439243? We flip the 819439243th bit. That's all we do. What do you do then? Well, you compare that new number you created, with the first number, the second number, and so on. And you see that this number is different to all the other numbers. What does it mean? That it doesn't matter what your enumeration scheme is, there is at least one number that will always be out of your enumeration scheme, and so your enumeration scheme will fail (so this is an absurd also, as we were supposing that this specific enumeration scheme -that could be any enum scheme- was complete). So you don't have as many integers as reals, so QED. Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity. As you see, it doens't matter, as in this proof you never need to do the sum of all numbers.
    - Retric 18y ago